Rearrangement?

[Gareth56]

I just can't see this rearrangment, could someone please explain it?

 

Thanks

 

F - Tcosx = 0

B - Wsinx = 0

 

all that boils down to tanX = B - W/F

 

I understand that cosX/sinX = tanX but it's the other bits I can't work out.

[Air]

I just can't see this rearrangment, could someone please explain it?

 

Thanks

 

F - Tcosx = 0

B - Wsinx = 0

 

all that boils down to tanX = B - W/F

 

I understand that cosX/sinX = tanX but it's the other bits I can't work out.

 

I don't see how you 'boiled' it down. I get:

 

[math]\frac{B}{W}=\sin x[/math]

[math]\frac{F}{T}=\cos x[/math]

 

[math]\frac{B}{W} \div \frac{F}{T} = \tan x[/math]

[math]\frac{B}{W} \times \frac{T}{F} = \tan x[/math]

[math]\frac{BT}{WF} = \tan x[/math]

[ajb]

Indeed Air, that is as far as I have got also.

 

How is T eliminated?

Edited September 8, 2008 by ajb

[DrP]

I just can't see this rearrangment, could someone please explain it?

 

Thanks

 

F - Tcosx = 0

B - Wsinx = 0

 

all that boils down to

 

I understand that cosX/sinX = tanX but it's the other bits I can't work out.

 

I get the same as Air (after I worked out that that Tan is Sin over Cos - not the other way round like you had it).

 

Is that tanX = B - W/F the answer you have been given and need to prove?

[Gareth56]

Sorry guys I made a typo, It should read:-

 

F - Tcosx = 0

B - W-Tsinx = 0

 

Apologies again:embarass:

[ajb]

You mean [math] \tan X = \frac{B-W}{F}[/math]?

[Gareth56]

You mean [math] \tan X = \frac{B-W}{F}[/math]?

 

Yes that's it. I don't know how to do that fancy equation thingy you do.

[the tree]

I don't know how to do that fancy equation thingy you do.

See the Quick LaTeX tutorial thread, read and inwardly digest.

[ajb]

All you have to do is take the ratio and the answer is right there. Write it as F = stuff and B-W = stuff and divide to get tan.

[Gareth56]

OK, So I have F = TcosX and B-W = TsinX ; what I'm uncertain about is the mechanics of bringing them together.

 

I knew I should have listened in maths class!

[ajb]

[math]\Rightarrow \frac{B-W}{F} = \frac{T}{T}\frac{\sin X}{\cos X}= \tan X[/math]

 

[math]\because[/math]

 

[math]\frac{T}{T} = 1[/math]

 

and

 

[math]\tan X = \frac{\sin X}{\cos X}[/math]

[Gareth56]

Ok thanks but what I'm not seeing is how do you equate

 

[math]

\frac{B-W}{F} = \frac{T}{T}\frac{\sin X}{\cos X}

[/math]

 

When we initially had two separate equations. The problem is I don't know what you meant when you said "All you have to do is take the ratio"

 

(Really sorry to appear thick here)

[ajb]

Assuming that nothing here is zero, you can simply divide one function by another. That is all I have done.

[Gareth56]

Function ??? I though they were 2 equations.

[DrP]

Look at ajb's post #11. The T/T go to 1 and the sin/cos go to Tan.

[Gareth56]

My problem is (other than being thick in this respect) is how does:

 

F - Tcosx = 0

 

B - W-Tsinx = 0

 

become:

 

[math]

\frac{B-W}{F} = \frac{T}{T}\frac{\sin X}{\cos X}

[/math]

 

I've heard words like divide and ratios but I just cannot see how you divide 2 equations.

 

I can get as far as:

 

F = Tcosx

 

and

 

B - W = Tsinx

 

But it's the steps that lead to:

 

[math]

\frac{B-W}{F} = \frac{T}{T}\frac{\sin X}{\cos X}

[/math]

 

that have me stumped.

[DrP]

OK - your perfectly entitled to do the following:

 

IF:

A=B and C=D

 

Then:

 

A/C = B/D

 

you are dividing one equation over the other.

 

So with:

F = Tcosx (1)

 

and

 

B - W = Tsinx (2)

 

you get F / (B-W) = Tcosx / Tsinx by dividing equ. (1) over equ. (2)

[Gareth56]

Oh I see many thanks for the clarification. I suppose to close the matter I'm bound to ask "why are you entitled to do such an operation"? Or again is it glaringly obvious to all by the uninitiated

[DrP]

Try it: Substitute some numbers in for A, B, C, and D. example - A and B = 3 and C and D = 6. so we have A=B and C=D. So A/C = B/D = 3/6 = 3/6. and so on for whatever A,B,C and D are.

 

In your case A = F, B = Tsin, C = B-W and D = Tcosx.

 

 

Trig questions do get ALOT harder I am afraid!!

[Bignose]

Oh I see many thanks for the clarification. I suppose to close the matter I'm bound to ask "why are you entitled to do such an operation"? Or again is it glaringly obvious to all by the uninitiated

 

Let's do it in a few separate steps:

 

Let's start with an equation:

 

y = z (call this equation 1)

 

(I'm going to use different letters, so that you see it in a bunch of different ways.)

 

And, given another equation

 

h=r (call this equation 2)

 

Now, let's divide both sides of equation 1 by something, namely h.

 

y/h = z/h. (equation 3)

 

Right? since you divided both side by the exact same thing, this is OK (assuming h doesn't equal 0).

 

Now, since we know from equation 2 that h=r, we can replace any h with an r and still be completely right. So, let's replace one of the h's in (3) with an r.

 

y/h = z/r (equation 4)

 

Final result.. it looks like you divided equation (1) by equation (2), but all the step are valid so long as the denominators aren't zero.